Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$ 
Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$

It is pretty easy to see that our goal is to choose from each multiplier in this polynomial $x,y,z,1$ in certain amounts. Since sum of powers equals $12$ and we have $16$ multipliers, we have to choose four $1$'s. There are $\displaystyle {8 \choose 4}$ ways to choose $1$ as part of our product (actually, four $1$'s). But how do we handle what has left? It seems like in such solution we have to consider many cases.
 A: Objective: To find coefficient of $\ x^3y^4z^5\ $ in $\ (x + y + z)^8(x + y + z + 1)^8$
First expand $$(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^r$$
Now multiplying $(1+(x+y+z))^8$ with $(x+y+z)^8$ gives us,
$$(x+y+z)^8(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^{r+8} \label{1} \tag{1}$$
Now using multinomial theorem, coefficient of $x^3y^4z^5$ in $(x+y+z)^n$ is $\Large{}{n \choose {3,\ 4,\ 5}}$ only when $n=12$.
Applying the same logic on equation $(1)$, we get the coefficient as
$$\sum_{r=0}^{8} {8 \choose r}* {8+r \choose {3,\ 4,\ 5}}$$ so we can fix $r=4$ and the answer we get is $${8\choose4}*{12\choose{3,4,5}}$$
For computationally rigorous answer see Wolfram Alpha coefficient finder
A: You want to pick 3 $x$'s, 4 $y$'s, and 5 $z$'s from the remaining 12 'multipliers', so you are now literally picking which multiplier you want for each variable. Hence there are
$$\binom{12}{3}\binom{12-3}{4}\binom{12-3-4}{5}$$
ways to do so
$\textbf{Side note}$: the expression here are often written as
$$\binom{12}{3,4,5}=\frac{12!}{3!4!5!}$$
as a multinomial coefficient. The order 3,4,5 does not matter (for a simple proof, write the binomial coefs in terms of factorials, and things will cancel out nicely)
